Let $g(x)=-(x+1)(x-1)^2$. For what value of $x$ does $g$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $1$ (Choice C) C $-\dfrac{1}{3}$ (Choice D) D $\dfrac{1}{3}$
Explanation: We can find the relative extrema (i.e. minima and maxima) of $g$ by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $g$ is $g'(x)=(1-x)(3x+1)$. $g'(x)=0$ for $x=1,-\dfrac{1}{3}$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-\dfrac{1}{3}$ and $x=1$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $x<-\frac{1}{3}$ $-\frac{1}{3}<x<1$ $-\frac{1}{3}$ $x>1$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<-\dfrac{1}{3}$ $x=-1$ $g'(-1)=-4<0$ $g$ is decreasing $\searrow$ $-\dfrac{1}{3}<x<1$ $x=0$ $g'\left(0\right)=1>0$ $g$ is increasing $\nearrow$ $x>1$ $x=2$ $g'(2)=-7<0$ $g$ is decreasing $\searrow$ Now let's look at the critical points: $x$ Before After Verdict $-\dfrac{1}{3}$ $\searrow$ $\nearrow$ Minimum $1$ $\nearrow$ $\searrow$ Maximum Now we can see that $g$ has a relative minimum at $x=-\dfrac{1}{3}$.